Rút gọn biểu thức:
a, \(\frac{-3}{4}\sqrt{9-4\sqrt{5}}\sqrt{\left(-8\right)^2\left(2+\sqrt{5}\right)^2}\)
b, \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
Rút gọn các biểu thức :
a) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
b) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
c)\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
Rút gọn biểu thức
a) \(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\)
b) \(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\)
a)
\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)
\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))
\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)
b)
\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)
\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))
\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)
rút gọn các biểu thức sau:
a,\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b,\(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
c,\(\sqrt{2+\sqrt{5-\sqrt{13-\sqrt{48}}}}\)
d,\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
1, Rút gọn biểu thức: \(A=\dfrac{-3}{4}.\sqrt{9-4\sqrt{5}}.\sqrt{\left(-8\right)^2.\left(2+\sqrt{5}\right)^2}\)
2, Với \(x=\sqrt{4+2\sqrt{3}}\). Tính giá trị biểu thức \(P=x^2-2x+2020\)
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
tiính hoặc rút gọn các biểu thức
a) \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}\)
b) \(\dfrac{\sqrt{4}}{\sqrt{5}}\)
c) \(\left(\sqrt[3]{9}\right)^2\)
d) \(\sqrt[5]{\sqrt{a}}\)
e) \(\sqrt[3]{2^6}\)
a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)
b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)
d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)
e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)
rút gọn biểu thức
a) \(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{9+2\sqrt{14}}\right)\)
b) \(\sqrt{\sqrt{13}-\sqrt{3-\sqrt{13}}-4\sqrt{3}}\)
c) \(\sqrt{80-\sqrt{321-16\sqrt{5}}-\sqrt{226-80\sqrt{5}-\sqrt{89-25\sqrt{5}}}}\)
d) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
e) \(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
f) \(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3}+\sqrt{5}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
g) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
Cho biểu thức: \(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a)Rút gọn biểu thức
b)Tính P với \(x=\frac{\sqrt{4+2\sqrt{3}}\left(\sqrt{x}-1\right)}{\sqrt{6+2\sqrt{5}-\sqrt{5}}}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
Rút gọn biểu thức \({\left[ {{{\left( {\frac{1}{3}} \right)}^2}} \right]^{\frac{1}{4}}}.{\left( {\sqrt 3 } \right)^5}\), ta được
A. \(\sqrt 3 \).
B. \(3\sqrt 3 \).
C. \(\frac{1}{{\sqrt 3 }}\).
D. 9.
\({\left[ {{{\left( {\frac{1}{3}} \right)}^2}} \right]^{\frac{1}{4}}}.{\left( {\sqrt 3 } \right)^5} = {\left( {\frac{1}{3}} \right)^{2.\frac{1}{4}}}.{\left( {{3^{\frac{1}{2}}}} \right)^5} = {\left( {{3^{ - 1}}} \right)^{\frac{1}{2}}}{.3^{\frac{1}{2}.5}} = {3^{ - \frac{1}{2}}}{.3^{\frac{5}{2}}} = {3^{ - \frac{1}{2} + \frac{5}{2}}} = {3^2} = 9\)
Chọn D.
Rút gọn các biểu thức sau:
a.\(\sqrt{5.\left(1-2\right)^2}\)
b.\(\sqrt{27.\left(2-\sqrt{5}\right)^2}\)
c.\(\sqrt{\frac{2}{\left(3-10\right)^2}}\)
d.\(\sqrt{\frac{5.\left(1-\sqrt{3}\right)^2}{4}}\)
e.\(\sqrt{\frac{4}{9-4\sqrt{5}}}+\sqrt{\frac{9}{9+4\sqrt{5}}}\)